How do you find #dy/dx# by implicit differentiation given #x^4+y^4=5#?

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Answer 1 · 2016-12-20T14:43:47

The answer is #=-x^3(5-x^4)^(-3/4)#

Let's do the differentiation

#x^4+y^4=5#

#4x^3+4y^3dy/dx=0#

#dy/dx=(-4x^3)/(4y^3)#

#=-x^3/y^3#

But, #y^4=5-x^4#

#y^3=(5-x^4)^(3/4)#

Therefore,

#dy/dx=-x^3/(5-x^4)^(3/4)=-x^3(5-x^4)^(-3/4)#